chapter 31:radioactivityand nuclear...
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College Physics Student Solutions Manual Chapter 31
219
CHAPTER 31: RADIOACTIVITY AND NUCLEAR PHYSICS 31.2 RADIATION DETECTION AND DETECTORS
1. The energy of 30.0 eV is required to ionize a molecule of the gas inside a Geiger tube, thereby producing an ion pair. Suppose a particle of ionizing radiation deposits 0.500 MeV of energy in this Geiger tube. What maximum number of ion pairs can it create?
Solution To calculate the number of pairs created, simply divide the total energy by energy
needed per pair: ( )( ) pairs 101.67eV/pair 30.0
eV/MeV 101.00MeV 0.500 pairs# 46
×=×
= . This is
the maximum number of ion pairs because it assumes all the energy goes to creating ion pairs and that there are no energy losses.
31.3 SUBSTRUCTURE OF THE NUCLEUS
9. (a) Calculate the radius of Ni58 , one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of Ni58 to that of Ha258 , one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom.
Solution (a) Using the equation 31
0Arr = we can approximate the radius of Ni58 :
( )( ) fm 4.6m 106.458m 102.1 1531153
1Ni0Ni =×=×== −−Arr
(b) Again using this equation this time we can approximate the radius of Ha258 :
( )( ) .fm 7.6m 106.7258m 102.1 153115
Ha =×=×= −−r
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Finally, taking the ratio of Ni to Ha gives: 1 to61.0m 10639.7m 10645.4
rr
15
15
Ha
Ni =××
=−
−
15. What is the ratio of the velocity of a 5.00-‐MeV β ray to that of an α particle with the same kinetic energy? This should confirm that β s travel much faster than α s even when relativity is taken into consideration. (See also Exercise 31.11.)
Solution We know that the kinetic energy for a relativistic particle is given by the equation
2rel )1(KE mc−= γ , and that since
21
2
2
1−
⎟⎟⎠
⎞⎜⎜⎝
⎛−=cv
γ , we can get an expression for the
speed ⎟⎟⎠
⎞⎜⎜⎝
⎛−= 22
2 11γc
v , or 2
11γ
−= cv
For the β particle: )MeV 511.0)(1(MeV 00.5KE −== γ , so that 785.9)1( =−γ or
785.10=γ . Thus, the velocity for the β particle is:
sm 10985.2)785.10(
11)sm 10998.2(11 82
82 ×=−×=−=
γβ cv
For the α particle: 22
uMeV 5.931)u 0026.4)(1(MeV 00.5KE cc
⎟⎠
⎞⎜⎝
⎛−== γ so that
00134.1=γ . Thus, the velocity for the α particle is:
sm 10551.1)00134.1(
11)sm 10998.2( 72
8 ×=−×=αv . Finally, the ratio of the
velocities is given by: 1 to3.19sm 1055.1sm 10985.2
7
8
=××
=α
β
vv
.
In other words, when the β andα particles have the same kinetic energy , the β particle is approximately 19 times faster than the α particle.
31.4 NUCLEAR DECAY AND CONSERVATION LAWS
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22. Electron capture by In106 .
Solution Referring to the electron capture equation, eNA
ZNAZ vYeX +→+ +−
−11 , we need to
calculate the values of Z and N . From the periodic table we know that indium has 49=Z and the element with 48=Z is cadmium. Using the equation ZNA += we
know that 5749106 === -A-ZN for indium and 58=N for cadmium. Putting this all together gives eve +→+ −
581064857
10649 CdIn
28. α decay producing Pb208 . The parent nuclide is in the decay series produced by Th232 , the only naturally occurring isotope of thorium.
Solution Since we know that Pb208 is the product of an α decay, 2422
42 He+→ −
−− NAZN
AZ YX tells
us that 2084 =−A , and since 822 =−Z from the periodic table, we then know that 126.822082 =−=−N So for the parent nucleus we have 84212 == , Z A and
N 128= . Therefore from the periodic table the parent nucleus is Po212 and the decay is 2
42126
20882128
21284 HePbPo +→
34. A rare decay mode has been observed in which Ra222 emits a C14 nucleus. (a) The decay equation is
€
222Ra→A X +14 C. Identify the nuclide XA . (b) Find the energy emitted in the decay. The mass of Ra222 is 222.015353 u.
Solution (a) The decay is 8146134
22288 CXRa +→ N
AZ , so we know that:
82688 208;14222 =−==−= ZA and 126,82208 =−=−= ZAN so from the
periodic table the element is lead and 12620882PbX=
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(b)
( ) ( ) ( )
( ) MeV 05.33uMeV/ 5.931u 105485.3
u105485.3u 003241.14u 976627.207u 015353.222
CPbRa
22
22
28
146126
20882134
22288
=⎟⎟⎠
⎞⎜⎜⎝
⎛×=Δ=
×=−−=
−−=Δ
−
−
ccmcE
mmmm
40. (a) Write the complete +β decay equation for C11 . (b) Calculate the energy released in
the decay. The masses of C11 and B11 are 11.011433 and 11.009305 u, respectively.
Solution (a) Using eNA
ZNAZ vYX ++→ +
+− β11 and the periodic table we can get the complete
decay equation: ev++→ +β61155
116 BC
(b) To calculate the energy emitted we first need to calculate the change in mass. The change in mass is the mass of the parent minus the mass of the daughter and the positron it created. The mass given for the parent and the daughter, however, are given for the neutral atoms. So the carbon has one additional electron than the boron and we must subtract an additional mass of the electron to get the correct change in mass.
( ) ( )[ ]( )[ ] u 10031.1u 00054858.02u 009305.11u 011433.11
2BC3
1111
−×=+−=
+−=Δ emmmm
( ) MeV 9602.0uMeV/ 5.931u 10031.1 2
232 =⎟⎟
⎠
⎞⎜⎜⎝
⎛×=Δ= − ccmcE
31.5 HALF-‐LIFE AND ACTIVITY
46. (a) Calculate the activity R in curies of 1.00 g of Ra226 . (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.
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Solution (a) First we must determine the number of atoms for radium. We use the molar mass
of 226 g/mol to get: ( ) atoms 106646.2mol
atoms 10022.6g 226
molg 1.00 2123
×=×
⎟⎟⎠
⎞⎜⎜⎝
⎛=N
Then using the equation 21
6930tN.R = , where we know the half life of Ra226 is
y 106.1 3× ,
( )( )
Ci 988.0Bq 1070.3
Ci
Bq 1066.3s 10156.3
y 1y 106.1
106646.20.693
16
1073
21
=⎟⎟⎠
⎞⎜⎜⎝
⎛
×
×=⎟⎠
⎞⎜⎝
⎛×
××
×=R
(b) The half life of Ra226 is more accurately known than it was when the Ci unit was established.
52. V50 has one of the longest known radioactive half-‐lives. In a difficult experiment, a researcher found that the activity of 1.00 kg of V50 is 1.75 Bq. What is the half-‐life in years?
Solution Using the equation
21
693.0tNR = , we can write the activity in terms of the half-‐life,
the molar mass, M , and the mass of the sample, m : [ ]
21
23
21
/)molatoms 1002.6()693.0(6930t
mMtN.R ×==
From the periodic table, g/mol 94.50=M , so
y 1048.1s 10156.3
y 1s 10681.4
)Bq 75.1)(molg94.50()g 1000)(molatoms 1002.6)(693.0(
177
24
23
21
×=⎟⎠
⎞⎜⎝
⎛×
×=
×=t
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58. The −β particles emitted in the decay of H3 (tritium) interact with matter to create light in a glow-‐in-‐the-‐dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of H3 . (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?
Solution Using the equation
21
693.0tNR = , we can write the activity in terms of the half-‐life,
the atomic mass, M and the mass of the sample m :
(a) ( )( )
21
21
693.0693.0t
MmtNR == . The atomic mass of tritium (from Appendix A) is
atomkg 100082.5u 1
kg 106605.1u 016050.3 2727
−−
×=⎟⎟⎠
⎞⎜⎜⎝
⎛ ×=M , and the half-‐life is
12.33 y (from Appendix B), so we can determine the original mass of tritium:
693.021 MRt
m = or
mg 56.1kg 1056.1
ys 10156.3
CiBq 1070.3
693.0)kg 100082.5)(y 33.12)(Ci 0.15(
6
71027
=×=
⎟⎟⎠
⎞⎜⎜⎝
⎛ ×⎟⎟⎠
⎞⎜⎜⎝
⎛ ××=
−
− Mm
(b) Ci 3.11y 33.12
)y 00.5(693.0exp)Ci 0.15(693.0exp2
100 =⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛−=−= tt
RteRR λ
31.6 BINDING ENERGY
71. Bi209 is the heaviest stable nuclide, and its A/BE is low compared with medium-‐mass nuclides. Calculate A/BE , the binding energy per nucleon, for Bi209 and compare it with the approximate value obtained from the graph in Figure 31.27.
Solution Dividing ( ) ( ){ } 21 ]H[BE cXmNmZm An −+= by A gives the binding energy per
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nucleon: ( ) ( )[ ]A
cmNmZmA
n2
12620983
1 BiHBE −+= .
We know that 83=Z (from the periodic table), 126=−= ZAN and the mass of the Bi209 nuclide is 208.908374 u (from Appendix A) so that:
[ ]
nucleonMeV 848.7uMeV 5.931
209u 980374.208)u 008665.1(126)u 007825.1(83BE 2
2
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−+= cc
A
This binding energy per nucleon is approximately the value given in the graph.
76. Unreasonable Results A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together. (a) Find the binding energy. (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent?
Solution (a)
[ ] [ ]
MeV 315.9uMeV 5.931u 02733.2)u 008665.1(2)particle(2BE 2
22
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−= cccmmn
(b) The binding energy cannot be negative; the nucleons would not stay together.
(c) The particle cannot be made from two neutrons.
31.7 TUNNELING
78. Integrated Concepts A 2.00-‐T magnetic field is applied perpendicular to the path of charged particles in a bubble chamber. What is the radius of curvature of the path of a 10 MeV proton in this field? Neglect any slowing along its path.
Solution Using the equation qBmvr = , we can determine the radius of a moving charge in a
magnetic field. First, we need to determine the velocity of the proton. Since the
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energy of the proton (10.0 MeV) is substantially less than the rest mass energy of the
proton (938 MeV), we know the velocity is non-‐relativistic and that 2
21mvE = .
Therefore,
cm 8.22m 228.0)T 00.2)(C 10602.1(
)sm 10377.4)(kg 106726.1(
,So).sm 10998.2)(1460.0(cMeV 27.938
MeV 0.1022
19
727
721
2
21
=×
××==
×=⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅=⎟
⎠
⎞⎜⎝
⎛=
−
−
qBmvr
mEv